Hot answers tagged

5

In this case, the Fourier transform is not the simplest method. But, if this is asked , we have to use the Fourier transform. Notation of Fourier transform of $u(x,t)$ relatively to the variable $x$ : $$\mathscr{F}_x\left(u(x,t) \right)(\omega)=U(\omega,t)$$ $$\frac{\partial^2 u}{\partial x^2} = \frac{\partial^2 u}{\partial x \partial t}$$ ...


2

You can successfully apply the method of characteristic curves. In your case, you should consider straight lines in the direction of $(a,b)$. So let $\phi$ the straight line through say $(x_0,0)$ with direction $(a,b)$. Then \begin{equation} \phi(t) = (x_0,0) + t(a,b). \end{equation} Let $U(t)=u(\phi(t))$. Then by the chain rule, $U$ is differentiable and ...


2

For nonlinear equations, you can't do "eigenfunction expansions" because there is no superposition principle: a linear combination of solutions is not likely to be a solution. So you can't build up a solution from simpler solutions. About the only hope, unless you can somehow transform the problem to a linear one, is that you can solve the whole problem in ...


2

$$(u_x+bu)_y+a(u_x+bu)=0$$ Let $v=u_x+bu$,then solve two ODEs


2

As Ian wrote, $u$ being a weak solution of $u_x+u_y=0$ means that $$\int_{\mathbb{R}^2} u(x,y) (v_x(x,y) + v_y(x,y))\, dx dy = 0\tag{1}$$ for every test function $v$. This follows by formal integration by parts of $$\int_{\mathbb{R}^2} (u_x(x,y)+u_y(x,y)) v(x,y) dx dy = 0\tag{2}$$ If $u(x,y)=f(x-y)$, then integral in $(1)$ becomes as ...


2

I believe you want to calculate $f$ locally. So take a chart whose image is a ball which contains $0$ and identify it with a neighborhood of $x\in M$ you have: $f(x)=\int_0^t\omega_{tx}(x)dt$.


2

Separation of variables works on regions that are rectangular in some particular coordinate system. The underlying operator must be separable in that coordinate system. For the Laplacian, that generally means an orthogonal coordinate system such as spherical coordinates, cylindrical coordinates, elliptic coordinates, etc.. A rectangular region in spherical ...


1

If $u_1, u_2$ are solutions then we know $a{[u_1]}_t+b{[u_1]}_x=0$ and $a{[u_2]}_t+b{[u_2]}_x=0$ Let $u=c_1u_1 + c_2u_2$.   So if $u$ actually is a solution then $a{\big[c_1u_1+c_2u_2\big]}_t +b{\big[c_1u_1+c_2u_2\big]}_x =0$.   Can you show that this is so? In the second case, If $u_1, u_2$ are solutions then we know ...


1

First, why this expression for $c$? If I recall correctly then $$ c=\sqrt{\frac{E}{\rho}}\,. $$ Second, the eigenvalues you found are the eigenvalues of the operator $$ -\frac{d^2}{dx^2}, $$ and the eigenvalues are $\omega_n^2$, it is easier to deal with squares, hence to relate everything you need to find the eigenvalues of $-A$ (you use $A$ in two ...


1

This is all explained in Rudin Functional Analysis up to ch. 6. You might want to start with the Evans book on PDEs instead of with Lions' notes; the book demands fewer prerequisites, although you do need to know what compact support is (essentially that the function is zero outside a compact set) and what it means for a function $f$ to be square integrable ...


1

Unfortunately, you really need to compute, either integrating (you know that integrals along any path with same initial and final points are equal), or solving: $$ \frac{\partial f}{\partial x_1}=1, \quad \frac{\partial f}{\partial x_2}=\sin(x_3),\quad \frac{\partial f}{\partial x_3}=x_2\cos{x_3},. $$ The computational difficulty is the same with both ...


1

There is a general method for finding an antiderivative of a closed differential form defined on a star-shaped region (more generally, on some region $D$ that’s the image of a star-shaped region) that’s particularly simple in the case of a one-form. It’s basically the same as what’s described in Tsemo Aristide’s answer, elsewhere. Step 1: Replace $x^i$ ...


1

The given initial conditions are probably not enough: specifying some derivatives at just $2$ points is unlikely to be a boundary condition that guarantees uniqueness. You would probably need boundary conditions on at least an open subset of the boundary, like the conditions you are suggesting, and you may need more; consider the example of boundary-value ...


1

The notes that you are following should be corrected by erasing the initial fraction: $$ \frac{dE}{dt}(t) = \int_\Omega (u_tu_{tt} + c(x)^2 \nabla u \cdot \nabla u_t + q(x)uu_t)\, dx. $$ In particular, the derivative of $(u_t)^2$ is obtained exactly as you wrote. As for the middle term, you have instead $$ \frac{\partial}{\partial t}|\nabla u|^2 = ...


1

For $0 \le s \le t$, the selfadjointness of $\Delta$ gives \begin{align} \frac{\partial}{\partial s}(u(t-s),v(s)) & =-(u'(t-s),v(s))+(u(t-s),v'(s)) \\ & =-(\Delta u(t-s),v(s))+(u(t-s),\Delta v(s)) =0. \end{align} Therefore, $$ (u(t-s),v(s)|_{s=0} = (u(t-s),v(s))|_{s=t} \\ (u(t),v_0) ...


1

From my understanding, the f(x,y,z) term is designated the "source term," and represents the rate at which heat is emitted or absorbed. Because the Neumann condition dictates that the total heat flux is zero, as you mentioned, and the source term is zero as well, this means that the heat is being absorbed at the same rate it is being generated, so overall ...


1

The answer is no, unless there are additional assumptions. Suppose $f(0) = g(0) = 1$ and $c = 1/2$. Then at interior points the solution will first increase and then decrease. A sufficient condition for $u$ to be decreasing in $t$ everywhere is that $c \ge \max\{f(0), g(0)\}$. Proof if $f$ and $g$ are smooth: Consider $v = \frac{\partial u}{\partial t}$ ...


1

It needs to be complete, in the metric generated by the inner product. This answer assumes your working in an infinite dimensional Inner product space. The problem is that an infinite linear combination may no longer be in the space.


1

Since you already took a course in numerical analysis before, you should be able to follow Gilbert Strang's lectures Computational Science and Engineering I Mathematical Methods for Engineers II


1

With your hypotheses, and for the function $f(x)=x$, the Fourier series of sines is equal to $f$ everywhere, except at $x=\pi$, where it is $0$ (as you can check explicitly, at this point, simply substituting $x=\pi$ in the series that you already wrote). More generally, the Fourier series of sines of a function $f$ with the hypotheses that you described ...


1

You have $dy/dx = y'/x' = 1/2(-x/y-y/x)$. Define $x/y = z$, then $y=x/z$, $dy/dx = (z-xdz/dx)/z^2$ so we have the ODE: $(z-xdz/dx) = z^2/2(-z-1/z) = 1/2(-z^3-z)$ $$ dz/dx x = 3/2 z+1/2 z^3$$ From there you get: $$dz/(3/2z +1/2 z^3) = dx/x$$ which you can integrate and solve.


1

Such ansatz makes sure that your $A$ is still symmetric. If you just multiply by $M^{-1}$, you will not have a symmetric matrix on the right-hand side of the equation. Here's how the given ansatz works: \begin{align} M\frac{d^2}{dt}X & = KX \\ M^{1/2}\frac{d^2}{dt}\left(M^{1/2}X\right) & = KX \\ \frac{d^2}{dt}\left(M^{1/2}X\right) & = ...


1

The way I view the use of the FT using periodic BCs is that the problem is really about expressing the solution in terms of Fourier series with the "transform" being the Fourier coefficients. The periodic BC's imply that $$u(x,t) = \sum_{n=-\infty}^{\infty} c_n(t) e^{i n \pi x} $$ where $$f(x) = \sum_{n=-\infty}^{\infty} c_n(0) e^{i n \pi x} $$ Thus, ...



Only top voted, non community-wiki answers of a minimum length are eligible